If $\gcd (a,b)=1$, and $\gcd(a,c)=1$, then $\gcd (a,bc)=1$ Help proving this? I'm really confused how to go about it..
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4Try writing $1=ax+by ; 1=as+(bc)t$, multiplying the two and combining terms in a convenient way. – user99680 Feb 14 '14 at 02:55
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There are many ways to prove this, what we need to know is what tools you've been given to use. – vadim123 Feb 14 '14 at 02:55
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By user99680's hint, we apply the Fundamental Theorem of Arithmetic to prove this statement.
Assume that $\gcd(a,b) = 1$ and $\gcd(a,c) = 1$. That is: for some integers $s$ and $t$, $$as + bt = 1$$
and for some integers $u$ and $v$, $$au + cv = 1$$
Multiply both equations to get $$\begin{aligned} (as + bt)(au + cv) &= 1\\ as(au) + as(cv) + au(bt) + cv(bt) &= 1\\ a(asu + scv + ubt) + (bc)vt &= 1 \end{aligned}$$
Assign arbitrary constants for $(asu + scv + ubt)$ and $vt$, say $x$ and $y$. Thus, we have
$$ax + (bc)y = 1$$
Therefore $\gcd(a,bc) = 1$
NasuSama
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how do we know that those $x, y$ give the minimum linear combination? because they give 1? If for example we had $\gcd(a,b) = g \wedge \gcd(a,c) = 1$ would $\gcd(a, bc) = g$ with the same proof hold? – tonythestark Nov 21 '22 at 19:17